factoring

J

jshaziza

Junior Member
Joined
Jan 26, 2007
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102
3a^2+3a+18.

I used FOIL to solve this problem but ran into a snag.

3(a^2+a+6)

From there I tried everything I could think of but I couldn't find anything that would fit. Could someone please show me how to work this problem? Thanks.
 
That's as far as you can take it. The roots are complex.

The discriminant is negative.
 
I am positive Denis, otherwise I wouldn't need your help. But now what would I put as my answer.

I also have a word problem that is similar.

The square of a positive number is seven more than six times the positive number. Find the number.

I tried solving it like this: x^2=6x+7
x^2+6x+7=0
Now either this is the same kind of problem like the one I needed help on or I made a mistake and x^2 needs to be negative. Could you please tell me which? Thanks
 
You got it, except you have a 'bad sign'.

\(\displaystyle \L\\x^{2}-6x-7=0\)

As for the first problem, there are two non-real solutions. Just use the quadratic formula to find them.
 
jshaziza said:
"I am positive Denis, otherwise I wouldn't need your help. But now what would I put as my answer."

Use the quadratic formula.

"x^2=6x+7
x^2+6x+7=0"

Should be x^2 - 6x - 7 = 0 ; simply factor:
(x - 7)(x + 1) = 0
x = 7 or x = -1

Galactus?
Click to expand...
 
Alright I am confused now, do I use the Quadratic formula to solve 3a^2+3a+18 or do I put as my answer 3(a^2+a+6). It tells me to factor completely.
 
Of course it can factors using complex numbers:
\(\displaystyle \L
3a^2 + 3a + 18 = 3\left( {a + \frac{{1 - i\sqrt {23} }}{2}} \right)\left( {a + \frac{{1 + i\sqrt {23} }}{2}} \right).\)
 
For your first problem, which as I understand it was to factor completely

3x<SUP>2</SUP> + 3x + 18

you'd start by removing the commonn factor of 3:

3(x<SUP>2</SUP> + x + 6)

Now, as you've discovered already, the expression inside the parentheses cannot be factored over the real numbers. DON'T overcomplicate things! You're done with
3(x<SUP>2</SUP> + x + 6)

It isn't an equation....you don't need to use the quadratic formula.

For the second problem, you have

x<SUP>2</SUP> = 6x + 7

Subtract 6x and 7 from both sides so that you get 0 on the right side:

x<SUP>2</SUP> - 6x - 7 = 0

The left side factors easily.....
 
jshaziza said:
pka, your method wasn't in my lesson. So I don't think I can use it.
Click to expand...
So why would you you have given something that you can not use? I would complain bitterly! Someone owes you an explanation!
You need money back!
 
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